Finding a line tangent to two points of a graph WITHOUT calculus Link to picture of graph and question 
My friends and I have tried figuring out the equation using the hint and different forms of equations of lines (eg. point-slope, two-point, etc.) but we find that we always have too many unknowns. As well, I've tried using the fact that at the zeros for f(x), D(x)=L(x) but I couldn't figure anything out.
Does anyone have any insight on how to approach this problem? Thanks in advance!
 A: HINT: $D(x)$ is a quartic with two double roots, so it must have the form $D(x)=(x-r)^2(x-s)^2$, where $r$ and $s$ are the roots. On the other hand, if $L(x)=a+bx$, then
$$(x-r)^2(x-s)^2=a+bx-2x^2-x^3+x^4\;,$$
so
$$x^4-2(r+s)x^3+(r^2+4rs+s^2)x^2-2rs(r+s)x+r^2s^2=x^4-x^3-2x^2+bx+a\;.$$
Equate coefficients:
$$\begin{align*}
-2(r+s)&=-1\\
r^2+4rs+s^2&=-2\\
-2rs(r+s)&=b\\
r^2s^2&=a
\end{align*}$$
Note that you can solve for $r+s$. You can also rewrite the second equation as
$$(r+s)^2+2rs=-2\;.$$
Can you finish solving for $a$ and $b$ from here?
A: The function $D(x)$ must have 2 second order zeros , so you must be able to write
$$D(x)=a(x-b)^2(x-c)^2 \\ =ax^4-2a(b+c)x^3+ a(b^2+c^2 +4bc)x^2 +-2a(bc^2+b^2c)x + ab^2c^2$$ for real numbers $a,b,c$
from the definition of $D$ and using $L(x)=mx+d$
$$ D(x)= -x^2(x+1)(x-2) - (mx+d) 
\\= -x^4 +x^3 +2x^2 - mx-d$$
Equating coefficients of $x$ gives you 5 equations and 5 unknowns
$$\begin{eqnarray}  \\a &= -1
\\2b+2c&=1
\\(b+c)^2 + 2bc &=-2
\\2bc(b+c) &= -m
\\b^2c^2 &=d
   \end{eqnarray}   $$
